If `z_1=a + ib and z_2 = c + id` are complex numbers such that `|z_1|=|z_2|=1 and Re(z_1 bar z_2)=0` , then the pair ofcomplex nunmbers `omega=a+ic and omega_2=b+id` satisfies

Options

A. `|omega_(1)|=1`
B. `|omega_(2)|=1`
C. `Re(omega_(1)baromega_(2)) = 0`
D. `Im(omega_(1)baromega_(2))=0`

Answer

A, B, C

Explanation

`|Z_(1)|=|Z_(2)| = 1`
`rArr a^(2) + b^(2) =c^(2) + a^(2) = 1" "(1)`
and `Re(z_(1)barz_(2)) =0`
`rArr Re{(a+ib)(c+id)} = 0`
`rArrac + bd = 0" "(2)`
Now from Eqs. (1) and (2) we get
`a^(2) + b^(2) =1`
` rArr a^(2) + (a^(2)c^(2))/(d^(2)) = 1`
`rArr a^(2) =d^(2)" "(3)`
Also, `c^(2)+d^(2) =1`
`rArr c^(2) +(a^(2)+c^(2))/(b^(2))=1`
`rArr b^(2) = c^(2)`
`|omega_(1)| = sqrt(a^(2) + b^(2)) = sqrt(a^(2) +b^(2)) = 1` [From (1) and (4)]
`and |omega_(2)|=sqrt(b^(2) + d^(2))= sqrt(c^(2) + d^(2))=1`[From (1) and (4)]
Further `Re(omega_(1)baromega_(2)) = Re{(a+ ic)(b-id)}`
`ab+ cd`
`=ab -(ac^(2))/(b)" "["From (2)"]`
`= (ab^(2) - ac^(2))/(b) =0" "["Form (4)"]`
Also, `Im(omega_(1)baromega_(2)) = bc-ad`
`= bc-a(-(ac)/(b)) = ((a^(2) +b^(2))c)/(b) = (c)/(b) = pm1ne 0`
`therefore |oemga_(1)|= 1, |omega_(2)|=1 and Re(omega_(1)baromega_(2))=0`